1. **Problem Statement:** Evaluate exactly the sum $$\sum_{i=25}^{150} \left(\frac{1}{i+4} - \frac{1}{i+5}\right)$$ 2. **Formula and Important Rule:** This is a telescoping series where terms cancel out in sequence. The general idea is that many intermediate terms will cancel, leaving only a few terms from the start and end. 3. **Rewrite the sum:** $$\sum_{i=25}^{150} \left(\frac{1}{i+4} - \frac{1}{i+5}\right) = \sum_{i=25}^{150} \frac{1}{i+4} - \sum_{i=25}^{150} \frac{1}{i+5}$$ 4. **Change the index in the second sum:** Let $j = i + 1$, then when $i=25$, $j=26$, and when $i=150$, $j=151$. So, $$\sum_{i=25}^{150} \frac{1}{i+5} = \sum_{j=26}^{151} \frac{1}{j+4}$$ But since $j = i+1$, the second sum is: $$\sum_{j=26}^{151} \frac{1}{j+4}$$ 5. **Rewrite both sums explicitly:** $$\sum_{i=25}^{150} \frac{1}{i+4} = \frac{1}{29} + \frac{1}{30} + \cdots + \frac{1}{154}$$ $$\sum_{j=26}^{151} \frac{1}{j+4} = \frac{1}{30} + \frac{1}{31} + \cdots + \frac{1}{155}$$ 6. **Subtract the sums:** Most terms cancel out: $$\left(\frac{1}{29} + \frac{1}{30} + \cdots + \frac{1}{154}\right) - \left(\frac{1}{30} + \frac{1}{31} + \cdots + \frac{1}{155}\right) = \frac{1}{29} - \frac{1}{155}$$ 7. **Final evaluation:** $$\frac{1}{29} - \frac{1}{155} = \frac{155 - 29}{29 \times 155} = \frac{126}{4495}$$ **Answer:** $$\sum_{i=25}^{150} \left(\frac{1}{i+4} - \frac{1}{i+5}\right) = \frac{126}{4495}$$